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# Help pls!!!

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The numerator and denominator of a fraction are positive integers. If both of them are increased by 1, the value of the new fraction would be 0.1 greater than the value of the original fraction. Compute the number of different original fractions satisfying the above conditions.

I know the answer, but I don't know how to get to it quickly. Can someone help me?

Apr 20, 2022

#1
+2666
+1

Write as an equation: $${x \over y} + {1 \over 10} = { x+1 \over y+1}$$

Simplify this to: $${{10x+y} \over 10y} = {x+1\over y+1}$$

Cross multiply and simplify into: $$y^2-9y=-10x$$

Factor as: $$y(y-9)=-10x$$

Now, you need to manually check all cases where \(0 .

I don't think there is a faster way.

Apr 20, 2022
#2
+126978
+1

x / y   + 1/10  =   (  x + 1)   / (y + 1)

[ 10 x + y  ] / [ 10y ]  =  ( x + 1)  /( y + 1 )

[10x + y] (y + 1)  =  10y ( x + 1)

10xy + y^2 + 10x + y =  10xy + 10y

y^2 - 9y  =  -10x

y^2  - 9y  + 10x  = 0      set this  up as a  quadratic in y

Using the Q Formula

9  ±   sqrt  [  9^2  - 4 * 1 * 10x ]                           9     ±    sqrt [  81  - 40x ]

_____________________________    =        ______________________  =  y

2                                                        2

We see that since  x is positive......then  the  radical  will only be an integer when x =  2  which produces a perfect square  of 1

And  y  becomes  [  9  + 1 ]  / 2    =   5    or   y  =  [9 -1] / 2   = 4

Check

(2/4) + 1/10  =     3 / 5                 (2 / 5 )  + 1/10   =   3/6

(1/2) + 1/10 =    3/5                     25 / 50 =   1/2

(12) / 20  = 3/5   =  true                 1/2  = 1/2     =  true

The orignal fractions  are    (2/4)  = 1/2        and  (2/5)

Apr 20, 2022
edited by CPhill  Apr 20, 2022